Abaloon is ascending at a height of 100 m above the ground when a packet is dropped from the ballon. After how much time does it reach the ground? (g=10m/s^2)
answer is 5 second tell the explation
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Answer:
v=v0-gt=5–9.81t
s=s0+v0t-0.5gt^2
s=100+5t-0.5(9.81)t^2
when it reaches the ground s=0
9.81t^2–10t-200=0
t={-(-10)+/-[(-10)^2–4(9.81)(-200)]^0.5}/2(9.81)
={10+/-[100+7848]^0.5}/19.62
={10+/-[7948]^0.5}/19.62={10+/-89.15}/19.62=5.05 sec
Therefore it will take about 5.05 seconds to hit the ground.
s=s0+v0t-0.5gt^2
s=100+5t-0.5(9.81)t^2
when it reaches the ground s=0
9.81t^2–10t-200=0
t={-(-10)+/-[(-10)^2–4(9.81)(-200)]^0.5}/2(9.81)
={10+/-[100+7848]^0.5}/19.62
={10+/-[7948]^0.5}/19.62={10+/-89.15}/19.62=5.05 sec
only the positive root has meaning.
Therefore it will take about 5.05 seconds to hit the ground.
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